Answer:
The carbon atom of the \[\overset{+\delta
}{\mathop{CH}}\,=\overset{-\delta }{\mathop{O}}\,\] group carries a small
positive charge due to greater electro negativity of O over C. In other words,
C atom of the CHO group acts as an electrophile. Now electron-donating groups
such as \[C{{H}_{3}}\] and \[C{{H}_{3}}O\] decrease while electron-withdrawing
groups such as \[Cl\] and \[N{{O}_{2}}\] increase the electrophilicity of this
aldehydic group. Since \[OC{{H}_{3}}\] is a stronger electron donating group
than \[C{{H}_{3}}\] and\[N{{O}_{2}}\] is a stronger electron-withdrawing than
Cl, therefore, the electrophilicity of the four aldehydes decreases in the
order : \[II>III>I>IV.\]
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